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The Hodge star is given by $$*(dx^{i_1}\wedge dx^{i_2}\wedge....\wedge dx_{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2....i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}$$ The question is to prove $$**\omega_p=(-1)^{p(n-p)}\omega_p$$ by repeating the * operator on a p-form $\omega_p$. Now, how shall I prove it? Here, $e_{ijk....}$ is antisymmetric tensor in n-dimension

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    $\begingroup$ I think there's one extra "$*$" in the first equation. Also, what do you mean by $e_{\cdots}$? $\endgroup$ – hjhjhj57 Mar 2 '15 at 6:13
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First of all, note that in the formula you give us, $$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}e_{i_1 i_2\dots i_p i_{p+1}...i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}, $$ there is a summation involved! In fact, there are $n-p$ sums, since the indices are being contracted. I won't use Einstein's convention in this post.

Explicitly we have: $$*(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=\frac{1}{(n-p)!}\sum_{j_{p+1},\dots j_{n}=1}^ne_{i_1 i_2\dots i_p j_{p+1}...j_n}dx^{j_{p+1}}\wedge dx^{j_{p+2}}\wedge....\wedge dx^{j_n}, $$

where $e_{\cdots}$ is the Levi-Civita tensor.

In fact you're summing the same $(n-p)$-form $(n-p)!$ times! This follows from the definition of the Levi-Civita tensor and that $dx_i\wedge dx_j = -dx_j\wedge dx_i$.

To find how many times you're summing the same term observe that whenever an index is repeated, the term vanishes. In this case, the indices in the sum can take only $n-p$ distinct values and must all be different. So we get $(n-p)!$ possible combinations.

From this we get that (no sum involved!): $$ *(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p})=e_{i_1 \dots i_n}dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}. $$ Now, applying again the star operator we get: \begin{align} **(dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p}) &=e_{i_1 \dots i_n}*dx^{i_{p+1}}\wedge dx^{i_{p+2}}\wedge....\wedge dx^{i_n}\\ &=e_{i_1\dots i_p i_{p+1}\dots i_n} e_{i_{p+1}\dots i_{n} i_1\dots i_{p}} dx^{i_1}\wedge dx^{i_2}\wedge\dots\wedge dx^{i_p}. \end{align} To conclude, observe that $$ e_{i_1\dots i_p i_{p+1}\dots i_n} e_{i_{p+1}\dots i_{n} i_1\dots i_{p}} = (-1)^{p(n-p)}. $$ (To show this count how many permutations you must do on the second term to obtain the first.)

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  • $\begingroup$ As I have already mentioned, here 'e' represents antisymmetric tensor in n- dimension, I should have better used epsilon symbol instead of 'e' for representing tensor. Now, how to proceed from here. $\endgroup$ – Roshan Shrestha Mar 2 '15 at 7:01
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    $\begingroup$ @RoshanShrestha So you're using Einstein's convention and there is a summation involved, right? $\endgroup$ – hjhjhj57 Mar 2 '15 at 7:06
  • $\begingroup$ @ hjhjhj57 Sorry mate, but I am very new at this differential geometry, definitely have to review my earlier notes on tensor. I am as lost as ever. Any help would be very helpful. $\endgroup$ – Roshan Shrestha Mar 2 '15 at 7:08
  • $\begingroup$ @RoshanShrestha feel free to ask if something's not clear. $\endgroup$ – hjhjhj57 Mar 2 '15 at 8:29
  • $\begingroup$ Why we are not using (n-p)! in our proof ? $\endgroup$ – Roshan Shrestha Mar 2 '15 at 14:16

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